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3 years ago

Explain Newton’s 3 laws of motion by using the example of a rollercoaster?

1 answer:

5 0

Newton's First law states the Law of Inertia. The Law of Inertia says that an object at rest must stay at rest unless an applied force is acted upon it. Roller coasters run by this law, they must be pulled by a motor to get started. ... At ride will continue until another opposite force is applied to the ride, the breaks.

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The apple became weightless when it fell...

Tems11 [23]

The answer I believe would be a sorry if I’m wrong

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3 years ago

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In a college homecoming competition, eighteen students lift a sports car. While holding the car off the ground, each student exe

Nata [24]

Answer:

Explanation:

Given

Each student exert a force of $F=400 N$

Let mass of car be m

there are 18 students who lifts the car

Total force by 18 students $F=18\times 400=7200 N$

therefore weight of car $W=7200$

mass of car $m=\frac{W}{g}$

$m=\frac{7200}{9.8}=734.69 kg$

(b) $7200 N \approx 1618.624\ Pound-force$

$734.69 kg\approx 1619.71 Pounds$

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3 years ago

In the picture at left. The launch height is 10m ball B mass 5kg has a launch velocity of 12 m/s and ball A mass 8 kg has a laun

Hitman42 [59]

Answer:

Explanation:

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3 years ago

A 98 N ball is suspended from a cable so that it hangs 3.5 m above the earth. Find the mass of the ball and the

expeople1 [14]

Answer:

Yes

Explanation:

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3 years ago

An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7

Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

$v = \frac{I}{nqA}$

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

$n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}$

Now, we can find the drift speed:

$v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s$

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!

4 0

2 years ago